# On the Dirichlet beta function sum $\sum_{k=2}^\infty\Big[1-\beta(k) \Big]$

Given the Dirichlet beta function,

$$\beta(k) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^k}$$

(The cases k = 2 is Catalan's constant.) It seems,

$$\sum_{k=2}^\infty\Big[1-\beta(k) \Big] = \frac{1}{4}\big(\pi+\log(4)-4\big)=0.131971\dots$$

or, in general, for some constant p > 0,

$$\sum_{k=2}^\infty\left[1-\sum_{n=0}^\infty\frac{(-1)^n}{(pn+1)^k} \right] = \sum_{m=1}^\infty\frac{1}{2p^2m^2+3pm+1}$$

Anyone knows how to prove the general proposed equality? (This is similar to the question on the zeta sum here.)

Here is a way to derive a slightly different looking result:

Notice that $$\sum_{k=2}^{\infty}\left[1-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(pn+1)^{k}}\right]=\sum_{k=2}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(pn+1)^{k}}$$

$$=\sum_{n=1}^{\infty}(-1)^{n-1}\sum_{k=2}^{\infty}\frac{1}{(pn+1)^{k}}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(pn+1)^{2}}\sum_{k=0}^{\infty}\frac{1}{(pn+1)^{k}}.$$ Now, since $$\sum_{k=0}^{\infty}\frac{1}{(pn+1)^{k}}=\frac{1}{1-\frac{1}{pn+1}}=\frac{pn+1}{pn},$$ our series is

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn(pn+1)}.$$

Plugging in the case $p=2$ seems to agree with your first identity.

Remark: Using partial fractions, we can go a bit further. Notice that $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn(pn+1)}=\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{pn}-\frac{1}{pn+1}\right)=\frac{\log 2}{p}-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn+1}$$

Suppose $p$ is an integer, and let $\zeta_{p}$ be a $p^{th}$ root of unity. Then consider $$\frac{\log\left(1+z\right)}{z}+\frac{\log\left(1+\zeta_{p}z\right)}{\zeta_{p}z}+\cdots+\frac{\log\left(1+\zeta_{p}^{p-1}z\right)}{\zeta_{p}^{p-1}z}=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}z^{n-1}\sum_{k=0}^{p-1}\zeta_{p}^{k(n-1)}$$

$$=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{pn+1}z^{pn}.$$ Letting $z=1,$ we have the identity $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{pn+1}=\sum_{k=0}^{p-1}\frac{\log\left(1+\zeta_{p}^{k}\right)}{\zeta_{p}^{k}z},$$ so our original series is $$\frac{1}{p}\log2+\sum_{k=0}^{p-1}\frac{\log\left(1+\zeta_{p}^{k}\right)}{\zeta_{p}^{k}z}.$$

• That was fast. :-) WolframAlpha says my non-alternating series and your alternating one are in fact equal. Commented Jun 24, 2012 at 16:44
• @Tito: Ahhh, good point. Commented Jun 24, 2012 at 17:11
• At the end, you said, "Letting $z=1$," but there's still a $z$ there. Is that a typo? Commented Aug 29, 2014 at 12:28

Let $$A(p,s)=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(pn+1)^{s}}, \quad p,s>0.$$

An integral representation for $$A(p,s)$$, is $$A(p,s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{1+e^{-px}} \, \mathrm dx.$$

The proof is straightforward:

\begin{align} \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{1+e^{-px}} \, \mathrm dx &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} x^{s-1} e^{-x} \sum_{n=0}^{\infty} (-1)^{n}e^{-pnx} \, \mathrm dx \\ &= \frac{1}{\Gamma(s)} \sum_{n=0}^{\infty} (-1)^{n} \int_{0}^{\infty} x^{s-1}e^{-(pn+1)x} \, \mathrm dx \\ &= \frac{1}{\Gamma(s)} \sum_{n=0}^{\infty} (-1)^{n} \frac{\Gamma(s)}{(pn+1)^{s}} \\ &= A(p,s). \end{align}

Therefore, an integral representation for $$1- A(p,s)$$ is \begin{align} 1- A(p,s) &= \frac{1}{\Gamma(s)}\int_{0}^{\infty} x^{s-1} e^{-x} \, \mathrm dx - \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{1+e^{-px}} \, \mathrm dx \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1} e^{-(p+1)x} }{1+e^{-px}} \, \mathrm dx. \end{align}

Using this representation, we have \begin{align} \sum_{k=1}^{\infty} \left(1-A(p,k) \right) &= \sum_{k=1}^{\infty} \frac{1}{\Gamma(k)} \int_{0}^{\infty} \frac{x^{k-1}e^{-(p+1)x}}{1+e^{-px}} \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{e^{-(p+1)x}}{1+e^{-px}} \sum_{k=1}^{\infty} \frac{x^{k-1}}{\Gamma(k)} \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{e^{-px}}{1+e^{-px}} \, \mathrm dx \\ &= \frac{1}{p} \int_{0}^{1} \frac{1}{1+u} \, \mathrm du \\ &= \frac{\ln(2)}{p}. \end{align}

If we start the summation from $$k=2$$ instead, we then have $$\sum_{k=2}^{\infty} \left(1-A(p,k) \right) = \frac{\ln(2)}{p} - \left(1-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{pn+1} \right).$$

The value of $$\sum_{n=0}^{\infty} \frac{(-1)^{n}}{pn+1}$$ can be expressed in terms of the digamma function. See equation (6) here.

Similarly, $$A^{*}(p,s) = \sum_{n=0}^{\infty} \frac{1}{(pn+1)^{s}}, \quad ( p>0, \color{red}{s>1}),$$ has the integral representation $$A^{*}(p,s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{1\color{red}{-}e^{-px}} \, \mathrm dx.$$

And an integral representation for $$1-A^{*}(p,s)$$ is $$A^{*}(p,s) = -\frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1} e^{-(p+1)x} }{1-e^{-px}} \, \mathrm dx.$$

Therefore, \begin{align} \sum_{k=\color{red}{2}}^{\infty} \left(1-A^{*}(p,k) \right) &= - \sum_{k=2}^{\infty} \frac{1}{\Gamma(k)} \int_{0}^{\infty} \frac{x^{k-1}e^{-(p+1)x}}{1-e^{-px}} \, \mathrm dx \\ &= -\int_{0}^{\infty} \frac{\left(e^{x}-1 \right)e^{-(p+1)x}}{1-e^{-px}} \, \mathrm dx \\ &= -\frac{1}{p} \int_{0}^{1} \frac{1-u^{1/p}}{1-u} \, \mathrm du \\ &= -\frac{H_{1/p} }{p} , \end{align}

where $$H_{x}$$ is the extension of the harmonic numbers to the real numbers.